# Recognizing a product of two 3d rotations (matrices)

1. Jun 14, 2008

### AA1983

Hi, I have a problem identifying some 3d rotation matrices. Actually I don't know if the result can be brought on the desired form, however it would make sense from a physics point of view. My two questions are given at the bottom.

$$\mathbf{s}=\left( \begin{array}{c} s_{x} \\ s_{y} \\ s_{z} \end{array} \right),\; \mathbf{S}=\left( \begin{array}{c} S_{x} \\ S_{y} \\ S_{z} \end{array} \right)$$

$$H=\left[\left( \begin{array}{ccc} 1-\left(1-C_{z}^{2}+C_{x}^{2}\right)\gamma^{2}& 0 & -2\gamma\left(1-C_{x}C_{z}\gamma\right) \\ 0 & 1-\left(1+C_{z}^{2}+C_{x}^{2}\right)\gamma^{2} & 0 \\ 2\gamma\left(1-C_{x}C_{z}\gamma\right)& 0 & 1-\left(1+C_{z}^{2}-C_{x}^{2}\right)\gamma^{2} \end{array} \right) \mathbf{s}\right]\cdot\mathbf{S}$$

(this describes Kondo effect in a quantum dot with spin-orbit interaction)

The goal is to bring H on a the form $$H = \left(\mathbf{A}\mathbf{s}\right)\cdot\left(\mathbf{B}\mathbf{S}\right)$$ where A and B are matrices describing rotations.

For Cx=Cz=0 :

$$H=\left[\left(1-\gamma^{2}\right)\underbrace{\left( \begin{array}{ccc} 1 & 0 & \theta \\ 0 & 1 & 0 \\ -\theta & 0 & 1 \end{array} \right)}_{R_{y}(\theta)+O(\theta^{2})} \mathbf{s}\right]\cdot\mathbf{S} \;\; \approx \;\; \left(1-\gamma^{2}\right)\left(R_{y}(\theta) \mathbf{s} \right)\cdot \mathbf{S} \;, \qquad \theta=\frac{-2\gamma}{1-\gamma^{2}}$$

That is, H is a vector product between spin S and a spin s that has been rotated around the y-axis with angle theta.

For Cx and Cz different from 0 :

In this case the matrix cannot be direct identified as a rotation around the x,y or z axis. The questions are now:

1) can it be a rotation around another axis?
2) Is it possible to write it as $$H = \left(\mathbf{A}\mathbf{s}\right)\cdot\left(\mathbf{B}\mathbf{S}\right)$$ where A and B are matrices describing rotations?

I would be glad if anybody has an idea about how to deal with this.

2. Jun 14, 2008

### lbrits

You essentially have written that $$H = \mathbf{s}_1^T \mathbf{D} \mathbf{s}_2$$, and the question is whether you can write $$\mathbf{D} = \mathbf{A} \mathbf{B}$$, where A and B are rotation matrices. Now, rotation matrices form a group, hence $$\mathbf{D}$$ should also be a rotation matrix. Is $$\mathbf{D}$$ orthogonal? I'm assuming it is.

Anyway, there isn't a unique way of writing D as a product of two rotations. One is to take A = D and B = identity. Maybe I've misunderstood.

3. Jun 16, 2008

### schieghoven

One of the real eigenvectors of the matrix is the y unit vector, so if it is a rotation matrix it can only possibly be a rotation about the y-axis. If it is a rotation matrix, the corresponding eigenvalue
$$1-\left(1+C_{z}^{2}+C_{x}^{2}\right)\gamma^{2}$$,
must be equal to 1. This can only happen if \gamma = 0. So except in this presumably trivial case, no, it is not a rotation matrix.

Dave

Last edited: Jun 16, 2008